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Let $X$ be a random (symmetric) matrix drawn from an unknown distribution. I have an estimate of $\lambda_{\min}(\mathbf{E}[X])$. Specifically, I have $$\lambda_{\min}(\mathbf{E}[X]) \geq c$$ a constact c which is very large .

Can we say anything about $$\mathbf{E}[\lambda_{\min}(X)]$$ I know that from Jensen's Inequality about concavity, that $$\mathbf{E}[\lambda_{\min}(X)] \leq \lambda_{\min}(\mathbf{E}[X])$$

But that doesnt help me in saying $$\mathbf{E}[\lambda_{\min}(X)] \geq c$$ My question is can the Jensen gap be so bad in such a situation that $\mathbf{E}[\lambda_{\min}(X)]$ is close to $0$ wheras $\lambda_{\min}(\mathbf{E}[X])$ is arbitrarily very large?

Rodrigo de Azevedo
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rostader
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1 Answers1

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To construct a counterexample, it will suffice to take diagonal matrices, where $\lambda_{min}(X)$ is just the smallest diagonal value.

Take $X_1,X_2$ independent uniform random variables. Define $$X=\begin{bmatrix}X_1&0\\0&X_2 \end{bmatrix}-\frac{5}{12}I.$$

We have that $\lambda_{min}(\mathbb{E}[X])=\lambda_{min}((1/2-5/12)I)=1/12$. On the other have $\lambda_{min}(X)=\min(X_1,X_2)-5/12.$ As $\mathbb{E}[\min(X_1,X_2)]=1/3$, we see that $\mathbb{E}[\lambda_{min}(X)]=-1/12$.

Now taking $c>0$, and replacing $X$ with $12cX$, we have that $\lambda_{min}(\mathbb{E}[X])=c$ and $c=-c$.

Thus $\lambda_{min}(\mathbb{E}[X])$ can be arbitrarily big, while $\lambda_{min}(\mathbb{E}[X])$ is arbitrarily small.

Pax
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